summation - Power Sum of Integers and Relationship with Sum of Squares and Sum of Cubes

Let $\displaystyle\sigma_m=\sum_{r=1}^n r^m$.

Refer to the tabulation of the power sum of integers here.

It is interesting to note that

$$\begin{align} \color{green}{\sigma_1}\ &=\frac 12 n(n+1)\\ \color{blue}{\sigma_2}\ &=\frac 16 n(n+1)(2n+1)\\ \color{red}{\sigma_3}\ &=\frac 14 n^2(n+1)^2&&=\color{green}{\sigma_1}^2\\ \sigma_4\ &=\frac 1{30}n(n+1)(2n+1)(3n^2+3n-1)&&=\frac 15\; \color{blue}{\sigma_2} \ (3n^2+3n-1)\\ \sigma_5\ &=\frac 1{12}n^2(n+1)^2(2n^2+2n-1)&&=\frac 13\; \color{red}{\sigma_3}\ (2n^2+2n-1)\\ \sigma_6\ &=\frac 1{42}n(n+1)(2n+1)(3n^4+6n^3-3n+1)&&=\frac 17\;\color{blue}{\sigma_2}\ (3n^4+6n^3-3n+1)\\ \sigma_7\ &=\frac 1{24}n^2(n+1)^2 (\cdots)&&=\frac 16\; \color{red}{\sigma_3}\ (\cdots)\\ \sigma_8\ &=\frac 1{90}n(n+1)(2n+1)(\cdots)&&=\frac 1{15}\color{blue}{\sigma_2}\ (\cdots)\\ \sigma_9\ &=\frac 1{20}n^2(n+1)^2(n^2+n-1)(\cdots)&&=\frac 15\; \color{red}{\sigma_3}\ (n^2+n-1)(\cdots)\\ \sigma_{10}&=\frac 1{66}n(n+1)(2n+1)(n^2+n-1)(\cdots)&&=\frac 1{11}\color{blue}{\sigma_2}\ (n^2+n-1)(\cdots) \end{align}$$
i.e.

• the sum of squares, $\sigma_2$, is a factor of sum of even powers greater than $2$, and


• the sum of cubes, $\sigma_3$, is a factor of sum of odd powers greater than $3$.

Is there a simple explanation for this, if possible without using Faulhaber's formula and Bernoulli numbers, etc?

and also,

Why does this occur only for $\sigma_2, \sigma_3$ but not for $\sigma_4, \sigma_5$, etc?